What is the new volume if 15.50 L of neon at 25.0 degree C is heated to 100.0 de

Question

What is the new volume if 15.50 L of neon at 25.0 degree C is heated to 100.0 degree C?
Show how a buffer made from formic acid (HCHO2) and formate ion (CHO2-) can neutralize added acid and added base? What is the new volume if 15.50 L of neon at 25.0 degree C is heated to 100.0 degree C?
Show how a buffer made from formic acid (HCHO2) and formate ion (CHO2-) can neutralize added acid and added base?
Show how a buffer made from formic acid (HCHO2) and formate ion (CHO2-) can neutralize added acid and added base?

Explanation / Answer

1.

V1 = 15.5 L T1 = 25 + 273 = 298 K

V2 = ? T2 = 100 + 273 = 373 K

Assuming P is constant

V1/T1 = V2/T2 (Charle's Law)

V2 = V1 * T1/T2

V2 = 15.5 * 298/373 = 12.38 L

So, V2 = 12.38 L

2.

A buffer made from formic acid (HCHO2) and formate ion (CHO2-) can neutralize added acid and added base.

If we add HCl then it effectively adds H+ ions to the mixture. Cl- ions do not enter into the net ionic equation.

The H+ ions will react with anion CHO2- . So H+ ions and CHO2- ions come together to form HCHO2. The net ionic reaction is then

CHO2 - (aq) + H+(aq) ---> HCHO2(aq)

Rather than changing the pH dramatically and making the solution acidic, the added H+ ions react to make molecules of a weak acid thus neutralizing the effect of added acid.

If we add NaOH effectively adds OH- ions to the mix. Na+ ions do not enter into the reaction.
The OH- ions will react with H+ ions, forming H2O. The OH- ions will also react with HCHO2 forming H2O and CHO2- ions. The net ionic reaction is then

HCHO2(aq) + OH-(aq) ---> H2O(l) + CHO2-(aq) hence it neutralises the added base by forming salt and water.

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